如果之前已经提出这个问题,我会道歉 - 我正在努力思考如何对我的搜索进行评论(因此这个尴尬的标题)!
我所拥有的是单字符值的数据框,如下所示:
-------------------------
| Parent | Daughter |
-------------------------
| A | B |
| B | C |
| B | D |
| A | E |
-------------------------
每个父母的位置总会有两个女儿(如完整的二叉树)。我试图编写一段代码,用于生成从顶级父级到最终子级的路径向量:
A B C
A B D
A E
但是父母的数量不同,载体的长度也各不相同。
我考虑过使用for循环,但是因为我觉得我需要每个级别都有一个问题。这棵树,我事先不知道。
我不一定想要代码,只是建议如何解决这个问题!但是,非常感谢任何帮助,谢谢!
编辑:我应该指出“从结束开始”'只是因为我认为这样会更容易 - 这当然不是必要的!
数据:
df <- data.frame(Parent = c("A", "B", "B", "A"), Daughter = c("B", "C", "D", "E"))
EDIT2:以下是所需结果的更多示例。如果我把桌子做得更大,那么:
-------------------------
| Parent | Daughter |
-------------------------
| A | B |
| B | C |
| B | D |
| A | E |
| C | F |
| C | G |
| E | H |
| E | I |
-------------------------
数据2:
df <- data.frame(Parent = c("A", "B", "B", "A", "C", "C", "E", "E"), Daughter = c("B", "C", "D", "E", "F", "G", "H", "I"))
然后我想要的载体是:
A B C F
A B C G
A B D
A E H
A E I
答案 0 :(得分:3)
使用 igraph 包,将数据帧转换为图形对象,获取路径,删除作为其他路径子集的路径。
library(igraph)
# example data
df <- data.frame(Parent = c("A", "B", "B", "A", "C", "C", "E", "E"),
Daughter = c("B", "C", "D", "E", "F", "G", "H", "I"))
# convert to graph object
g <- graph_from_data_frame(df)
# get all the paths, extract node ids from paths
res <- all_simple_paths(g, from = "A")
res <- lapply(res, as_ids)
# get index where vector is not subset of other vector
ix <- sapply(res, function(i) {
x <- sapply(res, function(j) length(intersect(i, j)))
sum(length(i) == x) == 1
})
# result
res <- res[ix]
# res
# [[1]]
# [1] "A" "B" "C" "F"
#
# [[2]]
# [1] "A" "B" "C" "G"
#
# [[3]]
# [1] "A" "B" "D"
#
# [[4]]
# [1] "A" "E" "H"
#
# [[5]]
# [1] "A" "E" "I"
答案 1 :(得分:1)
这可能会有所帮助:
parent <- "A"
lev <- df$Daughter[which(df$Parent == parent)]
output <- cbind(parent, lev)
while(length(lev) > 0){
lev <- df$Daughter[which(is.element(df$Parent, lev))]
output <- cbind(output, lev)
}
# which returns
> output
parent lev lev
[1,] "A" "B" "C"
[2,] "A" "E" "D"
这很容易转换为function(parent)
:
myfct <- function(parent){
lev <- df$Daughter[which(df$Parent == parent)]
output <- data.frame(parent, lev, stringsAsFactors = F)
while(length(lev) > 0){
dat <- df[which(is.element(df$Parent, lev)),]
newdat <- merge(x = output, y = dat, by.x = "lev", by.y = "Parent", all = TRUE)
col.first <- which(names(newdat) == "parent")
col.last <- which(names(newdat) == "Daughter")
col.sec.last <- which(names(newdat) == "lev")
col.rest <- setdiff(1:dim(newdat)[2], c(col.first, col.sec.last,col.last))
newdat <- newdat[, c(col.first, col.rest, col.sec.last, col.last)]
names(newdat)[2:(length(names(newdat))-1)] <- paste0("x.",2:(length(names(newdat))-1))
names(newdat)[length(names(newdat))] <- "lev"
output <- newdat
lev <- df$Daughter[which(is.element(df$Parent, lev))]
}
cols <- as.numeric(which(!sapply(output, function(x)all(is.na(x)))))
output <- output[,cols]
return(output)
}
这里可以应用这个功能:
parents.list <- unique(df$Parent)
sapply(parents.list, myfct)
# which returns
$A
parent x.2 x.3 x.4
1 A B C F
2 A B C G
3 A B D <NA>
4 A E H <NA>
5 A E I <NA>
$B
parent x.2 x.3
1 B C F
2 B C G
3 B D <NA>
$C
parent x.2
1 C F
2 C G
$E
parent x.2
1 E H
2 E I
现在您可以随时修改它以更改输出的结构。
修改强>
关键是添加while
。我编辑了我的代码,现在它应该工作而不必指定级别数。